Question

The solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What is the Ksp of barium carbonate

Answer 1

Ksp = 2.58x10⁻⁹

Step-by-step explanation:

The equilibrium that takes place is:

• BaCO₃(s) ↔ Ba⁺² + CO₃⁻³

• Ksp = [Ba⁺²] [CO₃⁻³]

The molar concentration of Ba⁺² at equilibrium is the solubility. The same can be said about CO₃⁻³:

• Ksp = s * s

• Ksp = s²

So now we can calculate the Ksp from the solubility, after converting the solubility from g/L to mol/L:

• 0.0100 g BaCO₃/L ÷ 197.34 g/mol = 5.07x10⁻⁵ mol/L

Ksp = (5.07x10⁻⁵ mol/L)²

• Ksp = 2.58x10⁻⁹